# The Formula

Let $0 \le a_{n}\leq b_{n}$ for all n
If $\sum\limits_{n=1}^{\infty}b_{n}$ converges, then $\sum\limits_{n=1}^{\infty}a_{n}$ also converges
If $\sum\limits_{n=1}^{\infty}a_{n}$ diverges, then $\sum\limits_{n=1}^{\infty}b_{n}$ also diverges

# Examples

If $\lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=\rho , 0<\rho<\infty$,
then both converge or both diverge .
If $\lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=0 \text{, and } \sum b_{n}$ converges
then $\sum a_{n}$ converges
If $\lim\limits_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=\infty \text{, and } \sum b_{n}$ diverges
Then $\sum a_{n}$ diverges.
To apply the Comparison Test to a series $\sum c_{n}$ you need to find another series $\sum u_{n}$ that will allow for a useful comparison.

# When should you use this test ???

• Case 1: If you suspect that $\sum c_{n}$ converges, you should look for a convergent series $\sum u_{n}$ that is larger then $\sum c_{n}$.
• Case 2: If you suspect that $\sum c_{n}$ diverges, you should look for a divergent series $\sum u_{n}$ that is smaller then $\sum c_{n}$
However. The Direct Comparison Test is most often used to compare series that are similar to a P-series or Geometric Series. Furthermore, when searching for comparison, we often take the given series and make a new series with only the highest degree of n in the numerator and denominator

# Examples

 Series Comparison Relation $\sum\limits_{n=1}^{\infty}\frac{5}{2+3n+n^4}$ $\sum\limits_{n=1}^{\infty}\frac{5}{n^4}$ $\frac{5}{2+3n+n^4} \leq \frac{5}{n^4}$
 Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{5}{n^4}$ converges so does $\sum\limits_{n=1}^{\infty}\frac{5}{2+3n+n^4}$
 Series Comparison Relation $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n^2-1}}$ $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ $\frac{1}{\sqrt{n^2-1}} \geq \frac{1}{n}$
 Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ diverges so does $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n^2-1}}$
 Series Comparison Relation $\sum\limits_{n=1}^{\infty}\frac{1}{3^{n}+2}$ $\sum\limits_{n=1}^{\infty}\frac{1}{3^{n}}$ $\frac{1}{3^{n}+2} \leq \frac{1}{3^{n}}$
 Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{1}{3^{n}}$ converges so does $\sum\limits_{n=1}^{\infty}\frac{1}{3^{n}+2}$
 Series Comparison Relation $\sum\limits_{n=1}^{\infty}\frac{\sqrt{n}}{n^2+1}$ $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$ $\frac{\sqrt{n}}{n^2+1} \leq \frac{1}{n^{\frac{3}{2}}}$
 Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$ converges so does $\sum\limits_{n=1}^{\infty}\frac{\sqrt{n}}{n^2+1}$
 Series Comparison Relation $\sum\limits_{n=1}^{\infty}\frac{1+ln(n)}{n}$ $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ $\frac{1+ln(n)}{n} \geq \frac{1}{n}$
 Conclusion: Since $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ diverges so does $\sum\limits_{n=1}^{\infty}\frac{1+ln(n)}{n}$
;